1985 AMC 8 Problem 8

Attempt Problem 8 of the 1985 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

If a=−2,a = -2, the largest number in the set
{−3a, 4a, 24a, a2, 1}\left\{ -3a,\ 4a,\ \dfrac{24}{a},\ a^2,\ 1 \right\}
is
−3a-3a
4a4a
24a\dfrac{24}{a}
a2a^2
11
Answer: A
Concepts:substitution
Small hint:
Substitute a=−2a = -2 into each expression
Big hint:
Check signs first: 4a4a and 24a\frac{24}{a} are negative, while −3a-3a and a2a^2 are positive
Solution:
Substituting a=−2a = -2 gives the values −3a=6,-3a = 6, 4a=−8,4a = -8, 24a=−12,\dfrac{24}{a} = -12, a2=4,a^2 = 4, and 1.1.
The largest of these is 6,6, which is −3a.-3a.
Thus, the correct answer is A .
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